Hey guys, I seem to be struggling be with conical pendulum question.
A ball of mass of 2.9kg swings on a string length of 2.2m and moves at a speed of 4.4m/s in a horizontal circular path.
Find the angle.
Thanks for the help.
(Btw you can find this question on http://www.physicseducation.com.au, really good website)
Sup skrt skrt!
All you have to do is substitute the values into this equation and rearrange for theta:
tan(theta)= v^2 / rg . In this question the mass is irrelevant 
You get an answer of 41.92 degrees! (please correct me if the answer is wrong because I couldn't find the solution in the website)
I actually think the problem is significantly more complicated than that, as we don't know the radius (the
length of the pendulum is given).
There are two forces acting on the ball: its weight and the tension up the string. Given that the path of the ball is horizontal circular motion, the ball is in equilibrium vertically. That is, the vertical component of the tension must be equal in magnitude to the weight force. In addition, the centripetal force is a result of the horizontal component of the tension force (see diagram below).
My solution contains some Specialist Maths ideas:
}&=mg\ \left(1\right)\\T\sin{\left(\theta\right)}&=\frac{mv^2}{r}\ \left(2\right)\\r&=l\sin{\left(\theta\right)}\ \left(3\right)\\\left(2\right)&\div\left(1\right):\\\tan{\left(\theta\right)}&=\frac{v^2}{rg}\ \left(4\right)\\\left(3\right)&\rightarrow\left(4\right):\\\tan{\left(\theta\right)}&=\frac{v^2}{gl\sin{\left(\theta\right)}}\\\tan{\left(\theta\right)}\sin{\left(\theta\right)}&=\frac{v^2}{gl}\\\frac{\sin^2{\left(\theta\right)}}{\cos{\left(\theta\right)}}&=\frac{v^2}{gl}\\\frac{1-\cos^2{\left(\theta\right)}}{\cos{\left(\theta\right)}}&=\frac{v^2}{gl}\\1-\cos^2{\left(\theta\right)}&=\frac{v^2}{gl}\cos{\left(\theta\right)}\\0&=\cos^2{\left(\theta\right)}+\frac{v^2}{gl}\cos{\left(\theta\right)}-1\\0&=\left[\cos^2{\left(\theta\right)}+\frac{v^2}{gl}\cos{\left(\theta\right)}+\left(\frac{v^2}{2gl}\right)^2\right]-\left(\frac{v^4}{4g^2l^2}\right)-1\\0&=\left[\cos{\left(\theta\right)}+\frac{v^2}{2gl}\right]^2-\frac{v^4}{4g^2l^2}-\frac{4g^2l^2}{4g^2l^2}\\\frac{v^4+4g^2l^2}{4g^2l^2}&=\left[\cos{\left(\theta\right)}+\frac{v^2}{2gl}\right]^2\\\frac{\pm\sqrt{v^4+4g^2l^2}}{2gl}&=\cos{\left(\theta\right)}+\frac{v^2}{2gl}\\\frac{-v^2\pm\sqrt{v^4+4g^2l^2}}{2gl}&=\cos{\left(\theta\right)}\end{align*})
Because \(4g^2l^2>0\), we know that \(v^4+4g^2l^2>v^4\), which means that \(\sqrt{v^4}=v^2<\sqrt{v^4+4g^2l^2}\).
Additionally, because \(\theta\in\left[0,\frac{\pi}{2}\right]\), we also know that \(\cos{\left(\theta\right)}\in\left[0,1\right]\). That is, \(\cos{\left(\theta\right)}\) is positive.
This means that:
}=\frac{-v^2+\sqrt{v^4+4g^2l^2}}{2gl}\\\theta=\cos^{-1}\left(\frac{-v^2+\sqrt{v^4+4g^2l^2}}{2gl}\right)\end{align*})
Finally, substituting in \(m=2.9, l=2.2, v=4.4\) and \(g=9.8\) gives \(\theta\approx 49.67^\circ\).
Home
Help
Search
Login
